= Solution
Integrate $e^{iz}/z$ around the positively oriented indented upper-half-plane contour. The outer semicircle runs from $R$ to $-R$; the inner semicircle runs clockwise from $-r$ to $r$, excluding zero. The integrand is holomorphic inside this region, so <Cauchy's integral theorem> gives total <integral> zero.
The outer arc tends to zero by part (a) with $f(z)=1/z$ and $\lambda=1$. On the inner arc $z=re^{i\theta}$, $\theta$ decreases from $\pi$ to zero, and
$$
\int_{\rm inner}\frac{e^{iz}}z\,dz=i\int_\pi^0e^{ire^{i\theta}}d\theta\longrightarrow-i\pi.
$$
The two real segments combine as
$$
\int_{-R}^{-r}\frac{e^{ix}}x\,dx+\int_r^R\frac{e^{ix}}x\,dx=2i\int_r^R\frac{\sin x}{x}\,dx.
$$
Passing to the limits gives $2iI-i\pi=0$, and hence the <Dirichlet integral>
$$
\boxed{\int_0^\infty\frac{\sin x}{x}\,dx=\frac\pi2.}
$$
This is an ordinary conditionally convergent improper <integral>: the integrand tends to one at zero, and integration by parts gives a tail bounded by $2/A$ beyond $A$. The clockwise indentation sign is essential.
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